\documentclass[UTF8]{ctexart}
\title{第一次作业}
\author {07323110王若莹}
\date{2024年10月6号}
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\newtheorem{deff}{定义}[section]%标题是Theorem，编号按section，是section的下一级
\newtheorem{thm}{定理}[section]
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\begin{document}
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\section{07323110}
\subsection{第一大题}
\subsubsection{第一题}
\begin{proof}
因为$\left \{ A_n \right \} $是单调不减事件列，所以$A_n\subset A_n+1$,\thickspace n=1,2,\thickspace...\\
\[\varliminf_{n \to \infty}A_n = \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty}A_k = \bigcup_{n=1}^{\infty}A_n\]\\
又因为\[\forall n \in \mathbb{N}, A_n \subset \bigcup_{n=1}^{\infty}A_n\]\\
所以\[\varlimsup_{n \rightarrow \infty}\subset \bigcup_{n=1}^{\infty}A_n=\varliminf_{n\to\infty}A_n=\lim_{n\to\infty}A_n\]
\end{proof}
\subsubsection{第二题}
\begin{proof}
因为$\left \{ A_n \right \} $是单调不增事件列，所以$A_n\supset A_n+1$,\thickspace n=1,2,\thickspace...\\
\[\varlimsup_{n \to \infty}A_n = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty}A_k = \bigcap_{n=1}^{\infty}A_n\]\\
因为\[\forall n \in \mathbb{N},A_n\supset \bigcap_{n=1}^{\infty}A_n\]\\
所以\[\varliminf_{n \to \infty}A_n \supset\bigcap_{n=1}^{\infty}A_n=\varlimsup_{n \to \infty}A_n\]
\end{proof}

\subsection{第二大题}
\subsubsection{第一题}
\begin{proof}
\[x \in A \Longleftrightarrow 1_A (\omega)=1\Longleftrightarrow x \in \{\omega : 1_A(\omega)=1\}\]
\end{proof}
\subsubsection{第二题}
\begin{proof}
(必要性)
\[\text{若}x \in A,则1_A(x)=1.\quad\text{因为}\quad 1_A(x)\leq 1_B(x).\quad \text{所以}\quad x\in B\]
所以\[\text{若}\quad x \in A,\text{则}\quad x \in B.\]
所以\[A \subset B\]\\
(充分性)
\[\text{若}\thickspace A \subset B, \text{若}\thickspace 1_A(x)=1,\text{则}\thickspace x\in A.\quad \because A \subset B \quad \therefore x \in B \quad \therefore 1_B(x)=1 \therefore 1_A(x)\leq 1_B(x)\]
\end{proof}

\subsubsection{第三题}
\begin{proof}
\[x \in A \land x \in B \thickspace\text{时}\]
\[\thickspace 1_A(x)=1 \quad 1_B(x)=1\]\\
此时\[x \in A \cap B , 1_{A \cap B} (x) =1\therefore 1_{A\cap B} = 1_A(x)\cdot 1_B(x)=1\]

\[x \in A \land x \notin B \thickspace\text{时}\]
\[\thickspace 1_A(x)=1 \quad 1_B(x)=0\]\\
此时\[x \notin A \cap B , 1_{A \cap B} (x) =0\therefore 1_{A\cap B} = 1_A(x)\cdot 1_B(x)=0\]

\[x \notin A \land x \in B \thickspace\text{时}\]
\[\thickspace 1_A(x)=0 \quad 1_B(x)=1\]\\
此时\[x \notin A \cap B , 1_{A \cap B} (x) =0\therefore 1_{A\cap B} = 1_A(x)\cdot 1_B(x)=0\]

\[x \notin A \land x \notin B \thickspace\text{时}\]
\[\thickspace 1_A(x)=0 \quad 1_B(x)=0\]\\
此时\[x \notin A \cap B , 1_{A \cap B} (x) =0\therefore 1_{A\cap B} = 1_A(x)\cdot 1_B(x)=0\]
综上，$1_{A\cap B}=1_A \cdot 1_B$
\end{proof}

\subsubsection{第四题}
\begin{proof}
\[x\in A \land x \in B\text{时}\]
\[1_A(x)=1,1_B(x)=1,1_{A\cap B}(x)=1,1_{A \cup B}(x)=1\]
\[\therefore 1_{A\cup B}(x)=1_A(x) + 1_B(x)-1_{A\cap B}(x)=1\]

\[x\in A \land x \notin B\text{时}\]
\[1_A(x)=1,1_B(x)=0,1_{A\cap B}(x)=0,1_{A \cup B}(x)=1\]
\[\therefore 1_{A\cup B}(x)=1_A(x) + 1_B(x)-1_{A\cap B}(x)=1\]

\[x\notin A \land x \in B\text{时}\]
\[1_A(x)=0,1_B(x)=1,1_{A\cap B}(x)=0,1_{A \cup B}(x)=1\]
\[\therefore 1_{A\cup B}(x)=1_A(x) + 1_B(x)-1_{A\cap B}(x)=1\]

\[x\notin A \land x \notin B\text{时}\]
\[1_A(x)=0,1_B(x)=0,1_{A\cap B}(x)=0,1_{A \cup B}(x)=0\]
\[\therefore 1_{A\cup B}(x)=1_A(x) + 1_B(x)-1_{A\cap B}(x)=0\]
综上，\(1_{A\cup B}(x)=1_A(x) + 1_B(x)-1_{A\cap B}(x)\)
\end{proof}

\subsection{第三大题}
\begin{proof}
\[\omega \in \varliminf_{n \to \infty}A_n \Longleftrightarrow \omega \in \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty}
A_k \Longleftrightarrow \]
\[\exists k_1 \in \mathbb{N^+}, \omega \in \bigcap{k=n}^{\infty}A_n \Longleftrightarrow \omega \text{属于}A_k \text{后所有的}A_n\]
即除去有限多个$A_n$外，$\omega$属于余下所有的$A_n$.
\end{proof}

\end{document}
